# Do we have Any idea why the uncertainty principle occurs?

1. Sep 3, 2011

### cdux

I understood the Delayed Eraser experiment, and the Eraser experiment and the basic double slit experiment.

And all I can conclude is, why? Why is there such a necessity? I now understand that it's most probably impossible to learn the 2 properties at the same time but I don't see any mechanism that directs it. Only that it exists.

I've heard of the idea that "the particle does not exist" and "the wave does not exist", "it only exists as a function". But that gets it to be more bizzare. And it doesn't explain its existence either.

2. Sep 3, 2011

### Clever-Name

This isn't a question anyone can answer. The HUP exists because observation tells us it exists. As far as I know there is no deeper motivation for the HUP.

3. Sep 3, 2011

### mathman

The uncertainty principle is built into quantum theory.

4. Sep 3, 2011

### tom.stoer

The uncertainty principle is a consequence of the Hilbert space structure, i.e. state vectors and operators acting on them. Of course now one could ask whether we have any idea why we need Hilbert spaces ...

5. Sep 3, 2011

### jeebs

I'm no expert on this at all but I thought that you could also do a bit of messing around with bra's & ket's etc. and use the Cauchy-Schwarz identity to derive the uncertainty principle that way?
I don't know how to do this off the top of my head but I remember seeing it in a lecture a while back. I'm going to look it up now actually.

6. Sep 3, 2011

### Staff: Mentor

If you accept that (a) particles have wavelike aspects to their behavior, (b) the momentum of a particle is inversely proportional to wavelength (and therefore directly proportional to frequency) via de Broglie's famous equation $p = h / \lambda$, and (c) these waves superpose (add) like other kinds of waves do, then the Heisenberg uncertainty principle is inescapable.

All waves that superpose follow a similar uncertainty principle. For example, an electrical signal pulse with a spatial "width" $\Delta x$ contains waves with a range of frequencies $\Delta \nu$, and the product $\Delta x \Delta \nu$ obeys a relationship that is similar to the HUP, but with a different constant.

7. Sep 5, 2011

### zonde

Maybe it helps to consider such a mechanism as an answer to this "why?" question:
No two particles can take exactly the same trajectory inside experimental setup.

I am not sure where it can lead so it would be nice to hear some comments.

8. Sep 6, 2011

### edpell

Why the HUP? I would say it is all in the de Broglie's famous equation $p = h / \lambda$. If you know where the test particle is with small uncertainty it has high momentum. When the test particle touches the particle under measure it gets a big wack. If the momentum is small the position uncertainty is large.

There are no point particles to use as test particles.

9. Sep 6, 2011

### raymo39

someone has previously mentioned this, but in less detail, the HUP does come out of the commutative nature of operators. this can be proven using Dirac notation and an understanding of not-so-elementary linear algebra. the case which everyone knows
delx.delp >= hbar/2 is just due to the expectation value of the commutivity of the x and p operators.
but this does lead to broader questions like why does Dirac notation make everything so pretty, and how does Hilbert space relate to everything...